The probability distribution of a random vari | vedclass

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(B) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X = x_i) = k + 2k + 3k + 4k = 10k = 1 \implies k = 0.1$.The mean $

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$2$

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(B) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X = x_i) = k + 2k + 3k + 4k = 10k = 1 \implies k = 0.1$.The mean $\mu = E(X) = \sum x_i P(X = x_i) = (3 	imes 0.1) + (5 	imes 0.2) + (7 	imes 0.3) + (9 	imes 0.4) = 0.3 + 1.0 + 2.1 + 3.6 = 7.0$.Now,$E(X^2) = \sum x_i^2 P(X = x_i) = (3^2 	imes 0.1) + (5^2 	imes 0.2) + (7^2 	imes 0.3) + (9^2 	imes 0.4) = (9 	imes 0.1) + (25 	imes 0.2) + (49 	imes 0.3) + (81 	imes 0.4) = 0.9 + 5.0 + 14.7 + 32.4 = 53.0$.The variance $\sigma^2 = E(X^2) - [E(X)]^2 = 53.0 - (7.0)^2 = 53.0 - 49.0 = 4.0$.The standard deviation $\sigma = \sqrt{\sigma^2} = \sqrt{4.0} = 2$.

$X = x_i$	$3$	$5$	$7$	$9$
$P(X = x_i)$	$k$	$2k$	$3k$	$4k$

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

The probability distribution of a random variable $X$ is given by the following table:
$X = x_i$ $3$ $5$ $7$ $9$
$P(X = x_i)$ $k$ $2k$ $3k$ $4k$

Then the standard deviation of $X$ is

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The probability distribution of a random variable $X$ is given by the following table:$X = x_i$$3$$5$$7$$9$$P(X = x_i)$$k$$2k$$3k$$4k$Then the standard deviation of $X$ is